Difference between revisions of "Curves in Space and Vector-Valued Functions"

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Before we start calculating, we need to introduce the arc length function.
 
Before we start calculating, we need to introduce the arc length function.
{{Calculus/Def|title=Arc length function|text=Suppose that a piecewise-smooth curve with a vector function <math>\mathbf{r}(t)=\langle f(t),g(t),h(t)\rangle,a \le t \le b</math>, and the curve is traversed exactly once as <math>t</math> increases from <math>a</math> to <math>b</math>. The arc length function <math>s(t)</math> is:
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:<math>s(t)=\int_a^t\sqrt{\biggl(\frac{dx}{du}\biggr)^2+\biggl(\frac{dy}{du}\biggr)^2+\biggl(\frac{dz}{du}\biggr)^2}du</math>}}
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Suppose that a piecewise-smooth curve with a vector function <math>\mathbf{r}(t)=\langle f(t),g(t),h(t)\rangle,a \le t \le b</math>, and the curve is traversed exactly once as <math>t</math> increases from <math>a</math> to <math>b</math>. The arc length function <math>s(t)</math> is:
 +
:<math>s(t)=\int_a^t\sqrt{\biggl(\frac{dx}{du}\biggr)^2+\biggl(\frac{dy}{du}\biggr)^2+\biggl(\frac{dz}{du}\biggr)^2}du</math>
 +
 
 
According to the definition, the arc length function for our curve <math>\mathbf{r}(t) </math> should be<blockquote><math>s(t)=\int_0^t|\mathbf{r}'(u)|du=\int_0^t\sqrt{(-\sin u)^2+(\cos u)^2+1^2}du=\sqrt{2}u\bigg]_0^t=\sqrt{2}t </math>
 
According to the definition, the arc length function for our curve <math>\mathbf{r}(t) </math> should be<blockquote><math>s(t)=\int_0^t|\mathbf{r}'(u)|du=\int_0^t\sqrt{(-\sin u)^2+(\cos u)^2+1^2}du=\sqrt{2}u\bigg]_0^t=\sqrt{2}t </math>
  
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\int f_1(t)dt \\ \int f_2(t)dt \\ \vdots \\ \int f_n(t)dt \\
 
\int f_1(t)dt \\ \int f_2(t)dt \\ \vdots \\ \int f_n(t)dt \\
 
\end{pmatrix}
 
\end{pmatrix}
+\mathbf{C} </math></blockquote>Then, the arc length will become:<blockquote><math>L=\int_a^b|\mathbf{r}'(t)|dt\quad </math>or <math>\quad L=\int_a^b \sqrt{[f_1'(t)]^2+[f_2'(t)]^2+\cdots+[f_n'(t)]^2}dt </math></blockquote>{{BookCat}}
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+\mathbf{C} </math></blockquote>Then, the arc length will become:<blockquote><math>L=\int_a^b|\mathbf{r}'(t)|dt\quad </math>or <math>\quad L=\int_a^b \sqrt{[f_1'(t)]^2+[f_2'(t)]^2+\cdots+[f_n'(t)]^2}dt </math></blockquote>
  
 
==Resources==
 
==Resources==
* [https://en.wikibooks.org/wiki/Calculus/Vector_Functions Vector Functions], WikiBooks: Calculus
 
 
* [https://math.libretexts.org/Bookshelves/Calculus/Book%3A_Calculus_(OpenStax)/13%3A_Vector-Valued_Functions/13.1%3A_Vector-Valued_Functions_and_Space_Curves Vector-Valued Functions and Space Curves], Mathematics LibreTexts
 
* [https://math.libretexts.org/Bookshelves/Calculus/Book%3A_Calculus_(OpenStax)/13%3A_Vector-Valued_Functions/13.1%3A_Vector-Valued_Functions_and_Space_Curves Vector-Valued Functions and Space Curves], Mathematics LibreTexts
 +
 +
== Licensing ==
 +
Content obtained and/or adapted from:
 +
* [https://en.wikibooks.org/wiki/Calculus/Vector_Functions Vector Functions, WikiBooks: Calculus] under a CC BY-SA license

Latest revision as of 12:53, 14 November 2021

For more on curves in space, see Equations of Lines, Planes and Surfaces in Space

Introduction

If we have a function , we say that 's image (the set - or some subset of ) is a curve in and is its parametrization.

Parameterizations are not necessarily unique - for example, such that is one parametrization of the unit circle, and such that is a whole family of parameterizations of that circle.

Collision and intersection points

Say we have two different curves. It may be important to consider

  • points the two curves share - where they intersect
  • intersections which occur for the same value of - where they collide.

Intersection points

Firstly, we have two parameterizations and , and we want to find out when they intersect, this means that we want to know when the function values of each parametrization are the same. This means that we need to solve

because we're seeking the function values independent of the times they intersect.

For example, if we have and , and we want to find intersection points:

with solutions

So, the two curves intersect at the points .

Collision points

However, if we want to know when the points "collide", with and , we need to know when both the function values and the times are the same, so we need to solve instead

For example, using the same functions as before, and , and we want to find collision points:

which gives solutions So the collision points are .

We may want to do this to actually model physical problems, such as in ballistics.

Intersection vector functions

We can also use vector functions to represent the curve of intersection of two surfaces. For example, we want to know the curve of intersection of the cylinder and the plane .

Vector functions rely on parameterizations, so we can rewrite the equation of the cylinder into: , where .


From the equation of the plane, we know that . Thus the corresponding vector equation is:


Limits and Continuity

The limit of a vector function is defined by taking the limits of its component functions.

There is a vector function . If exist, then

And the requirement for continuity is also simple:

A vector function is continuous at if .

Derivatives and Integrals

Differentiation

Recall that the first derivative of a scalar function is defined as:

The first derivative of a vector function is defined in much the same way:

We can use this definition to prove that the derivative of a vector function can be presented as the derivative of its component functions.

Thus, using the same method, we can derive the second derivative and so on.

Derivatives of a vector function

There is a vector function . The first derivative of this vector function is:

}}So the th-order derivative should look like this:

Differentiation rules

Just like real-valued functions, there are some differentiation rules in the world of vector functions. The factor that makes vector differentiation rules slight more complicated is the product rule because there are two kinds of multiplication in vectors: dot product and cross product.

Suppose are differentiable vector functions and is a real-valued function. Then

  1. (addition)
  2. (scalar multiplication)
  3. (dot product)
  4. (cross product)
  5. (chain rule)

Naturally, we will prove that those rules are correct. Let us assume that and .


Rule 1: the addition rule

Rule 2: the scalar multiplication rule

Rule 3: the dot product rule

Rule 4: the cross product rule

Rule 5: the chain rule

Angle between curves

Inner-product-angle.png

We can then formulate the concept of the angle between two curves by considering the angle between the two tangent vectors. If two curves, parametrized by and intersect at some point, which means that

the angle between these two curves at is the angle between the tangent vectors and , using the dot product, is given by

Integration

Similar to real-valued functions, the definite integral of a vector function is defined as:

We can extend the Fundamental Theorem of Calculus to continuous vector functions as follows:

For indefinite integrals, the definition is:

Arc length

We can deduce the length of a curve with parametric equations , to be:

Since vector functions are fundamentally parametric equations with directions, we can utilize the formula above into the length of a space curve.

If the curve has the vector equation , or, equivalently, the parametric equations , where are continuous, then the length of the curve from to is:

For those who prefer simplicity, the formula can be rewritten into:

or

Reparametrization

Suppose that there is a vector function which describes the displacement of a particle with respect to time and has the equation:

However, for some reason, we do not want to know the displacement of this particle with respect to time. Instead, we want to know its displacement with respect to its traveled distance () from in the direction of increasing . In order to do so, we need to find a way to describe time as a function of distance. In other words, we need to find . We can use the formula for the arc length to establish the relationship between time and distance because arc length in this case describes the distance traveled by the particle.


Before we start calculating, we need to introduce the arc length function.

Suppose that a piecewise-smooth curve with a vector function , and the curve is traversed exactly once as increases from to . The arc length function is:

According to the definition, the arc length function for our curve should be

Note that because the initial point corresponds to the parameter value . Since it is in the direction of increasing , the integration direction should be from to .

Then we substitute the original function and get the answer:

Reparametrization has important applications in real life because sometimes we want to know a value with respect to different variables. In this case, instead of describing the path of a particle with respect to time, we described its path with respect to its distance, which will be very useful in certain situations.

Curvature

Terminology

Before we start discussing curvature, there are some important vectors and concepts we need to be at least aware of.

The unit tangent vector

In the differentiation section of this chapter, we discussed the derivatives of a vector function. We know that at is tangent to the curve at . is called the tangent vector. The unit tangent vector, however, eliminates the aspect of magnitude because it is defined as:

As we can see, the magnitude of the unit tangent vector is always . We can imagine that as the displacement of a particle with respect to time. So, the unit tangent vector can be perceived as the direction of the velocity of the particle with respect to time. It can also be perceived as the direction of the tangential acceleration of the particle with respect to time. We will discuss motion in space in the next section, but this is a useful method to intuitively understand some vectors.

The unit normal vector

The unit normal vector is defined as

The unit normal is orthogonal to the unit tangent because since , we can get that:


This means that is orthogonal to . Therefore, is orthogonal to . We can imagine that the unit normal vector is the direction of the normal acceleration of the particle with respect to time.

The binormal vector

The binormal vector is defined as

The binormal vector is perpendicular to both the unit tangent and the unit normal because of the properties of the cross product. The magnitude of the binormal is always 1 because

The normal plane, osculating plane, and the osculating circle

  • The normal plane is the plane determined by the normal and binormal vectors . The normal plane consists of all lines that are orthogonal to the tangent vector .
  • The osculating plane is the plane determined by the unit tangent and unit normal . It is the plane that comes closest to containing the part of the curve near a point where .
  • The osculating circle is the circle that lies in the osculating plane towards the direction of with a radius (the inverse of the curvature, which we will immediately discuss after this). It best describes how the curve behaves near the point where because it shares the same tangent, normal, and curvature at that point.

These concepts are very important in the branch of differential geometry and in its applications to the motion of spacecraft.

Curvature

The curvature of a curve at a given point is a measure of how quickly the curve changes direction at that point. We define it to be the magnitude of the rate of change of the unit tangent with respect to arc length. We use arc length so that the curvature will be independent of parametrization.

Suppose that a space curve has vector function , unit tangent vector , and arc length . The curvature of this curve is: .

There are two other ways to express the curvature. We can express curvature in terms of instead of by utilizing the chain rule (recall that ):

The third way is more complicated to deduce, but it is often more convenient to apply because it only requires and its derivatives.

And now for the proof for this theorem:

According to the definition of the unit tangent vector, we know that . So the second derivative of should be:

Now we calculate .

And then we calculate .

We rearrange the equation into:

Since , we can substitute with and get:

Here is a little summary on ways to calculate the curvature.

Definition With parametrization with respect to In terms of and its derivatives

Motion in space

Velocity and acceleration

Remember in 2-dimensional calculus, we mentioned that a particle with displacement function has velocity and acceleration . In vector functions, the definition is basically the same. Suppose a particle moves through space so that its position vector at time is , its velocity function and acceleration function are:

and

Simply put: .


The speed of the particle ignores the direction. It is the magnitude of the velocity vector: . The distance traveled by the particle from is , which is also the formula for the arc length.


With the help of the Fundamental Theorem of Calculus, we can deduce the velocity function and position function when we know the particle's acceleration.

and

Tangential and normal acceleration

We can split the acceleration vector into two components: the tangential acceleration and the normal acceleration . The tangential acceleration faces the same direction as the unit tangent vector (), and the normal acceleration faces the same direction as the unit normal vector (). Since both and are unit vectors, the acceleration vector can be written as the sum of two vectors:

Our goal is to figure out how to describe the two components.

Recall that , thus

Now we differentiate both sides of the equation,

So we get .

Recall that , thus .

Recall that , thus .

We substitute for to yield:

That leaves us with:

and

Of course, it will be more convenient if those components can be written in terms if and its derivatives. Suppose that is the angle between and , then we can write like this:

To sum it up:

and

Expansion

We only discussed vector functions with three variables . How about expanding our understanding of vector functions into variables? Suppose we have a curve with vector function:

Limits

The limit of a vector function is defined as:

Differentiation and integration

The derivative of a vector function is defined as:

, thus

All differentiation rules apply.


We can expand integration into:

Then, the arc length will become:

or

Resources

Licensing

Content obtained and/or adapted from: